Suppose I have to functions $f(x)$ and $g(x)$ such that $$ f(x) \leq g(x) \quad \forall x. $$

For a distribution $\pi(x)$ on $x$, is it necessarily true that $$ E_\pi[f(x)] \leq E_\pi[g(x)]? $$

My thinking is this is true due to the fact that $$ E_\pi[f(x)] = \int f(x) \pi(x) dx \leq \int g(x) \pi(x) dx = E_\pi[g(x)], $$ given that $\pi(x)$ is a non-negative function.


$\forall x$ $f(x)\leq g(x)$

As $\pi(x)$ is a distribution, we have $\pi(x)\geq0$. By multiplying $\pi(x)$ to the inequality :

$\pi(x)f(x)\leq \pi(x)g(x)$ $\forall x$

if $f$ and $g$ are measurable we have:

we will use the linearity of the integral : $\int\pi(x)f(x)\leq \int\pi(x)g(x)$ $\forall x$

$E_{\pi}[f(x)]\leq E_{\pi}[g(x)]$ $\forall x$

| cite | improve this answer | |
  • $\begingroup$ Thanks, this seems to match with my thinking. I have not taken a measure theory course, how would I go about determining whether $f$ and $g$ are measurable? $\endgroup$ – MontCarlo Jan 17 '19 at 17:18
  • $\begingroup$ The measurability of a function is based on the concept of measurable space. In my opinion, there is no intuitive definition of measurable function. But for simplicity, make sure that they are continuous functions. NB: a non continuous fonction can be a mesurable function. The contuinity ensures the measurability $\endgroup$ – Abdoulaye NIANG Jan 18 '19 at 13:18

$E(g(X))-E(f(X)) = E(g(X)-f(X))= \int (g(x)-f(x))\,\pi(x)\, dx \geq 0$ (since the integrand is everywhere non-negative).

You can generalize the argument in similar fashion.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.